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our question 8 states that the maximum
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speed of a train is 98 km/
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hour it takes 10 hours to cover the
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distance of 500 kilm find the ratio of
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its average speed to maximum speed now
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let us look at the information given in
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the question it is given in the question
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that maximum speed is equals to 90 kilm
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per hour total distance covered is equal
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kilm time taken to cover this distance
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hours now in this question we do not
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need to change unit from kilometer per
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hour to m/s because all the quantities
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given in the question are in terms of
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kilometers and hours so we have to find
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the ratio of average speed of the train
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to the maximum speed of the train so we
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have information about maximum speed of
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the train what we have to find here is
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average speed and we know that average
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speed is equals to total distance
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divided by time taken to cover that
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distance so we know that total distance
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covered is equals to 500 km and time in
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which this distance was covered is
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equals to 10 hours so average speed is
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equals to 500 km / by 10 R is equal to
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50 km per hour now the
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ratio of average speed to maximum speed
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is equals to 50 is to 90 which is equal
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to 5 is 9 and this is is your required
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answer question n states that a car
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starts from rest and acquire a velocity
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acceleration and second part is we have
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to find distance traveled by car assume
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uniform now before solving the question
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note down the information given in the
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question so in the question it is stated
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that car starts from rest since car is
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starting from rest so its initial
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to0 and the car is acquiring a velocity
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of 54 km/ hour in 2 second so our final
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velocity is equals to
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hour and time given is T is equals to 2
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seconds now here in this case final
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velocity is in kilometer per hour and
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time is in seconds so what we have to do
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is we have to convert this final
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m/s so we already know that 1 kilom
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36 m per second so final velocity is
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15 m/ second now in part one of the
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question we have to find the
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acceleration so we know that
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acceleration a is equals to change in
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upon time which is equal to 15 - 0 /
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2 here 15 is final velocity and Zer is
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our initial velocity so we get
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m/s Square now in second part of the
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question we have to find the distance
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traveled by the car and we have to
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assume that motion of a car is uniform
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so uniform motion means rate of change
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of velocity remains constant or we can
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say that acceleration of motion remains
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constant and distance traveled can be
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find using the equation s equal to UT +
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t² so we are using this particular
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equation because we have knowledge of
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all these quantities involved in this
5:12
equation for example we have the
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knowledge that the initial
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velocity which is VI is equals to
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0 acceleration we have just calculated
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7.5 m/s square and we also know the time
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is equals to 2 second so putting in all
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these values in this equation we get S =
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7.5 * 4 on calculating this we get S is
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our question 10 states that an object
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dropped from a cliff Falls with a
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constant acceleration of 10 m/s squar
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speed 5 Second after it was
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dropped so let us first gather the
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information given in the question so in
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the question it is given that an object
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is dropped from Cliff so acceleration of
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the object after dropping is a =
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m/s² we have to find its a speed 5
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Second after it was dropped so our time
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to 5 Second and our initial velocity VI
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is equals to 0 because initially stone
6:54
is at rest so here we would use the
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equation V is equal to U +
7:11
velocity so final velocity or velocity
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Second is equals to 0 + acceleration is
7:24
10 m/s Square time T which is equal to
7:31
seconds so our final velocity is equals
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second now this question 11 is a
7:47
conceptual question you don't have to do
7:49
any calculation for this question so
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let's look at our question it states
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that a ball is thrown upwards and it
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goes to the height 100 m and comes down
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so what is the net displacement and what
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is net distance so let us assume a ball
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goes up to this point so this is the
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initial point from which the ball is
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thrown upwards and the question is
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saying it reaches a height of 100 m and
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then again come down to this point only
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so again this point is also our final
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point so as the ball comes down to its
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initial point so net displacement is
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equals to zero and net distance is equal
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to total distance covered by the ball
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which is 100 m in going upward Direction
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and 100 m is the distance traveled by
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the ball when it comes back to the
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initial point so total distance or net
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distance is equal to 100 m+ 100 m which
